Every finite-dimensional vector space is isomorphic to its dual.
However for an infinite-dimensional vector space $E$ over a field $K$ this is always false since its dual $E^\ast$ is a vector space of strictly larger dimension: $dim_KE \lt dim_K E^\ast $ (dimensions are cardinals of course). This is a non-trivial statement for which our friend Andrea Ferretti has given an astonishingly unexpected proof here. This implies for example that a vector space of countably infinite dimension over a field $K$, like the polynomial ring $K[X]$, cannot be the dual of any $K$-vector space whatsoever.

So an infinite-dimensional vector space is not isomorphic to its dual but it could be isomorphic to the dual of another vector space and my question is: which vector spaces are isomorphic to the dual of some other vector space and which are not?

In order to make the question a little more precise, let me remind you of an amazing theorem, ascribed by Jacobson (page 246) to Kaplanski and Erdős:

The Kaplanski-Erdős theorem : Let $K$ be a field and $E$ an infinite-dimensional $K$-vector space . Then for the dual $E^\ast$ of $E$ the formula $dim_K (E^\ast) = card (E^\ast)$ obtains.

So now I can ask

A precise question : Is there a converse to the Kaplanski-Erdős condition i.e. if a $K$- vector space $V$ (automatically infinite dimensional !) satisfies $dim_K (V) = card (V)$ , is it the dual of some other vector space : $V \simeq E^\ast$? For example, is
$\mathbb R ^{(\mathbb R)}$ - which satisfies the Kaplanski-Erdős condition ( cf. the "useful formula" below) - a dual?

A vague request : Could you please give "concrete" examples of duals and non-duals among infinite-dimensional vector spaces?

A useful formula : In this context we have the pleasant formula for the cardinality of an infinite-dimensional $K$-vector space $V$ ( for which you can find a proof by another of our friends, Todd Trimble, here )

Dear Mariano, we'll see in the answers what this really is (Kaplanski-Erdős involves hard linear algebra), but you are definitely right that I should add the tag "set-theory". I have just done it: thanks for your remark.
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Georges ElencwajgFeb 3 '11 at 9:04

Note that if $\mathrm{dim}_K(V) = \lambda \geq \omega$, then a $K$-v.s. $W$ is (isomorphic to) $V^{\ast}$ iff $|W| = |K|^{\lambda}$ iff $\mathrm{dim}_K(W) = |K|^{\lambda}$. So a $K$-v.s. is isomorphic to another infinite dimensional $K$-v.s. iff it's cardinality, or equivalently, its dimension over $K$, is an infinite power of $|K|$. Now you may ask, which cardinals are infinite powers of $|K|$? The short answer is, there's a few basic things we can say about cardinal exponentiation, but aside from those basic restrictions, any other possibility is consistent relative to ZFC.
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Amit Kumar GuptaFeb 3 '11 at 17:15

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For some basics of cardinal arithmetic, see chapter 5 in Jech's "Set Theory." Easton showed that for $\kappa$ regular, $2^{\kappa}$ can be anything of cofinality greater than $\kappa$. Chapter 15 in that same book deals with applications of forcing and covers Easton forcing. The function $\gimel (\kappa) = \kappa ^{\mathrm{cf}(\kappa)}$ plays an important role in cardinal arithmetic, and in light of Easton's theorem, it's of most interest in the case of singular $\kappa$. Abraham and Magidor's article "Cardinal Arithmetic" in "The Handbook of Set Theory" is a good reference.
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Amit Kumar GuptaFeb 3 '11 at 19:08

3 Answers
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The $\mathbf{R}$-vector space $\mathbf{R}^{(\mathbf{R})}$ has dimension $\operatorname{card} \mathbf{R}$ by definition, so it is isomorphic to $\mathbf{R}^{\mathbf{N}}$ (by the Erdős-Kaplansky theorem and because $\operatorname{card}(\mathbf{R}^{\mathbf{N}}) = \operatorname{card} \mathbf{R}$). So $\mathbf{R}^{(\mathbf{R})}$ is isomorphic to the dual of $\mathbf{R}[X]$.

In general, your precise question is equivalent to the following purely set-theoretical question (which seems difficult). By the useful formula, the identity $\operatorname{card} V = \operatorname{dim}_K V$ is equivalent to $\operatorname{card} K \leq \operatorname{dim}_K V$. Let $\kappa = \operatorname{card} K$ and $\lambda = \dim_K V$, and assume $\kappa \leq \lambda$. Does there always exist an infinite cardinal $\alpha$ such that $\kappa^\alpha = \lambda$? (here $\alpha$ is meant to be the dimension of the vector space whose dual is $V$). In general, Stephen's answer shows that there are counterexamples.

EDIT : in order to explain why the question is difficult, consider a field $K$ such that $\operatorname{card} K = \aleph_1$ (for example, one could take $K=\mathbf{Q}((T_i)_{i \in I})$ with $\operatorname{card} I =\aleph_1$). Take a $K$-vector space $V$ of dimension $\aleph_1$. Then $V$ is a dual if and only if there exists $\alpha \geq \aleph_0$ such that $\aleph_1 = (\aleph_1)^\alpha$, which amounts to say that $\aleph_1 = 2^{\aleph_0}$. In other words, $V$ is a dual if and only if the continuum hypothesis holds.

François, let me point out that (regardless of the continuum hypothesis or any such assumption) there are cardinals that are not of the form $\kappa^\alpha$ for any infinite $\kappa,\alpha$. For example, take $\beth_\omega$, the supremum of the sequence ${}|{\mathbb N}|,2^{|{\mathbb N}|},2^{2^{|{\mathbb N}|}},\dots$ This is because of a result of König that asserts that $\kappa^{{\rm cf}(\kappa)}>\kappa$ for any infinite $\kappa$. Here, ${\rm cf}$ is the cofinality function, en.wikipedia.org/wiki/Cofinality
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Andres CaicedoFeb 4 '11 at 7:41

If $V$ is a vector space of infinite dimension over a field $K$, then the dimension of its dual is given by $\dim(V^*)=|K|^{\dim(V)}$. This is essentially just a restatement of what you have called the Kaplanski-Erdős theorem, because elements of $V^*$ correspond to functions $B\to K$, where $B$ is a basis of $V$, so $\dim(V^*)=|V^*|=|K^B|=|K|^{\dim(V)}$.

To answer your "precise question": Let $V$ be a vector space of dimension $\aleph_0$ over $\mathbb{Q}$. Then $|V|=\dim(V)$, yet $V$ is not isomorphic to the dual of any vector space, as its dimension is not of the form $|\mathbb{Q}|^\lambda$ for any infinite cardinal $\lambda$. Moreover, the same is true if $\aleph_0$ is replaced by any other strong limit cardinal, such as $\beth_\omega$.

Note that $\mathbb{R}^{(\mathbb{R})}$ is isomorphic to the dual of $\mathbb{R}^{(\mathbb{N})}$.

Stephen: Your "the same is true if..." is not quite correct. For example, $\beth_{\omega_1}^{\aleph_0}=\beth_{\omega_1}$. Here, $\omega_1$ is the first uncountable ordinal. (Your remark is true with $\beth_\omega$ and even with $\beth_\alpha$ whenever $\alpha$ has cofinality $\omega$, for example.)
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Andres CaicedoFeb 4 '11 at 7:37

@Andres Caicedo - I'm not claiming that $\beth_{\omega_1}^{\aleph_0}>\beth_{\omega_1}$. I only need that $\beth_{\omega_1}\cdot{\aleph_0}=\beth_{\omega_1}$ (to get $|V|=\dim(V)$) and that there is no cardinal $\lambda$ such that $\aleph_0^\lambda=\beth_{\omega_1}$ - and these statements are true, and don't involve $\beth_{\omega_1}^{\aleph_0}$.
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Stephen SFeb 4 '11 at 9:19

Let $K$ be a field of some size $\kappa$ and let $V$ be a $K$-vector space of dimension $\lambda$. If $\kappa\leq\lambda$, then the dimension of the dual of $V$ is simply $2^\lambda$. This is implies that vector spaces of large dimensions are duals iff
their dimension (and hence their size) is a power of 2. I don't have the time right now to elaborate what happens if the dimension of $V$ is small relatively to the size of $K$.